Simplify and expand the following expression: $ \dfrac{2}{5t - 20}- \dfrac{2}{3t + 30}+ \dfrac{3t}{t^2 + 6t - 40} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{2}{5t - 20} = \dfrac{2}{5(t - 4)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{2}{3t + 30} = \dfrac{2}{3(t + 10)}$ We can factor the quadratic in the third term: $ \dfrac{3t}{t^2 + 6t - 40} = \dfrac{3t}{(t - 4)(t + 10)}$ Now we have: $ \dfrac{2}{5(t - 4)}- \dfrac{2}{3(t + 10)}+ \dfrac{3t}{(t - 4)(t + 10)} $ The least common multiple of the denominators is: $ 15(t - 4)(t + 10)$ In order to get the first term over $15(t - 4)(t + 10)$ , multiply by $\dfrac{3(t + 10)}{3(t + 10)}$ $ \dfrac{2}{5(t - 4)} \times \dfrac{3(t + 10)}{3(t + 10)} = \dfrac{6(t + 10)}{15(t - 4)(t + 10)} $ In order to get the second term over $15(t - 4)(t + 10)$ , multiply by $\dfrac{5(t - 4)}{5(t - 4)}$ $ \dfrac{2}{3(t + 10)} \times \dfrac{5(t - 4)}{5(t - 4)} = \dfrac{10(t - 4)}{15(t - 4)(t + 10)} $ In order to get the third term over $15(t - 4)(t + 10)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{3t}{(t - 4)(t + 10)} \times \dfrac{15}{15} = \dfrac{45t}{15(t - 4)(t + 10)} $ Now we have: $ \dfrac{6(t + 10)}{15(t - 4)(t + 10)} - \dfrac{10(t - 4)}{15(t - 4)(t + 10)} + \dfrac{45t}{15(t - 4)(t + 10)} $ $ = \dfrac{ 6(t + 10) - 10(t - 4) + 45t} {15(t - 4)(t + 10)} $ Expand: $ = \dfrac{6t + 60 - 10t + 40 + 45t}{15t^2 + 90t - 600} $ $ = \dfrac{41t + 100}{15t^2 + 90t - 600}$